Figure shows a block P of mass m resting on a horizontal smooth floor at a distance $l$ from a rigid wall. Block is pushed toward right by a distance $3l/2$  and released. When block passes from its mean position another block of mass $m_1$ is placed on it which sticks to it due to friction. Find the value of $m_1$ so that the combined block just avoids collision with the left wall.

So velocity of block when passing from its mean position is given as  $v=A\omega =\frac{3l}{2}\sqrt{\frac{k}{m}}[As\omega =\sqrt{\frac{k}{m}}]$
If mass $m_1$  is added to it and just after if velocity of combined block becomes, from momentum conservation we have $mv=(m+m_1)v_1$
or  $v_1=\frac{m}{(m+m_1)}\left(\frac{3l}{2}\sqrt{\frac{k}{m}}\right)$
If this is the velocity of combined block at mean position, it must be given as
$v_1=A{\omega }_1[Now{\omega }_1=\sqrt{\frac{k}{m_1+m_2}}]$
Where $A_{1}and{\omega }_1$ are the new amplitude and angular frequency of SHM of the block. It is given that combined block just reaches the left wall thus the new amplitude of oscillation must be l so we have
$$\begin{gathered} \frac{m}{(m_1+m_2)}.\frac{3l}{2}\sqrt{\frac{k}{m}}=l_1\sqrt{\frac{k}{m+m_1}} \\ or\frac{3\sqrt{m}}{2\sqrt{m+m_1}}=1 \\ or9m=4m+4m_1 \\ or{m}_1=\frac{5}{4}m \end{gathered}$$